1. 11.5 Mathematical Induction
1 Peter 4:10-11 "As each has received a gift, use it to serve
one another, as good stewards of God's varied grace: whoever
speaks, as one who speaks oracles of God; whoever serves, as
one who serves by the strength that God supplies—in order
that in everything God may be glorified through Jesus Christ.
To him belong glory and dominion forever and ever. Amen."

2. Deductive Reasoning is piecing together known
facts to determine a specific conclusion.

3. Deductive Reasoning is piecing together known
facts to determine a specific conclusion.
Proving two triangles congruent

4. Deductive Reasoning is piecing together known
facts to determine a specific conclusion.
Proving two triangles congruent
a→b
b→c
c→d
∴ a→d

5. Deductive Reasoning is piecing together known
facts to determine a specific conclusion.
Proving two triangles congruent
a→b
b→c
c→d
∴ a→d
Inductive Reasoning is observing patterns and
then making generalizations.

6. Let’s consider the sum of the first k odd positive integers.

7. Let’s consider the sum of the first k odd positive integers.
1 =1

8. Let’s consider the sum of the first k odd positive integers.
1 =1
1+ 3 =4

9. Let’s consider the sum of the first k odd positive integers.
1 =1
1+ 3 =4
1+ 3 + 5 =9

10. Let’s consider the sum of the first k odd positive integers.
1 =1
1+ 3 =4
1+ 3 + 5 =9
1+ 3 + 5 + 7 = 16

11. Let’s consider the sum of the first k odd positive integers.
1 =1
1+ 3 =4
1+ 3 + 5 =9
1+ 3 + 5 + 7 = 16
1+ 3 + 5 + 7 + 9 = 25

12. Let’s consider the sum of the first k odd positive integers.
1 =1
1+ 3 =4
1+ 3 + 5 =9
1+ 3 + 5 + 7 = 16
1+ 3 + 5 + 7 + 9 = 25
Using Inductive Reasoning, our conjecture is that the sum
of the the first k odd positive integers is k .
2

13. Let’s consider the sum of the first k odd positive integers.
1 =1
1+ 3 =4
1+ 3 + 5 =9
1+ 3 + 5 + 7 = 16
1+ 3 + 5 + 7 + 9 = 25
Using Inductive Reasoning, our conjecture is that the sum
of the the first k odd positive integers is k .
2
A Math Induction Proof is used to prove that our
conjecture is true.

14. Not all conjectures are true ...

15. Not all conjectures are true ...
It has rained for the past three days ...
therefore, it will rain every day from now on.

16. Not all conjectures are true ...
It has rained for the past three days ...
therefore, it will rain every day from now on.
A counterexample is one case where the conjecture is not
true ... and that one counterexample would be enough to
disprove the conjecture.

17. Math Induction Proof

18. Math Induction Proof
The concept:
Suppose that we can prove that when a statement is
true for a given case, then it is also true for the next
case as well ...

19. Math Induction Proof
The concept:
Suppose that we can prove that when a statement is
true for a given case, then it is also true for the next
case as well ... If p ( k ) is true
then p ( k + 1) is true

20. Math Induction Proof
The concept:
Suppose that we can prove that when a statement is
true for a given case, then it is also true for the next
case as well ... If p ( k ) is true
then p ( k + 1) is true
then all we need do is show that
p (1) is true

21. Math Induction Proof
The concept:
Suppose that we can prove that when a statement is
true for a given case, then it is also true for the next
case as well ... If p ( k ) is true
then p ( k + 1) is true
then all we need do is show that
p (1) is true
It’s just like knocking dominos down ... set them all up so
that any given domino will knock down the next one ...
then knock over the first one! They all fall ...

22. Math Induction Proof
The concept:
Suppose that we can prove that when a statement is
true for a given case, then it is also true for the next
case as well ... If p ( k ) is true
then p ( k + 1) is true
then all we need do is show that
p (1) is true
It’s just like knocking dominos down ... set them all up so
that any given domino will knock down the next one ...
then knock over the first one! They all fall ...
The next slide states this formally.

23. The Principle of Math Induction

24. The Principle of Math Induction
For each natural number n , let p ( n ) be a statement
depending on n . Suppose that the following two
conditions are satisfied:
1. p (1) is true
2. For every natural number k,
if p ( k ) is true then p ( k + 1) is true.
Then p ( n ) is true for all natural numbers n .

25. Prove that the sum of the the first k odd positive
integers is k .
2

26. Prove that the sum of the the first k odd positive
integers is k .
2
2
1. p (1) : 1 = 1
1 = 1 ∴ p (1) is true

27. Prove that the sum of the the first k odd positive
integers is k .
2
2
2. p ( k ) : 1+ 3 + 5 +K + ( 2k − 1) = k

28. Prove that the sum of the the first k odd positive
integers is k .
2
2
2. p ( k ) : 1+ 3 + 5 +K + ( 2k − 1) = k
We assume p(k) to be true ...
Write p(k+1), then start with p(k) and use math
manipulation to get p(k) to become p(k+1).

29. Prove that the sum of the the first k odd positive
integers is k .
2
2
2. p ( k ) : 1+ 3 + 5 +K + ( 2k − 1) = k
We assume p(k) to be true ...
Write p(k+1), then start with p(k) and use math
manipulation to get p(k) to become p(k+1).
2
p ( k + 1) : 1+ 3 + 5 +K + ( 2k − 1) + ( 2k + 1) = ( k + 1)

30. Prove that the sum of the the first k odd positive
integers is k .
2
2
2. p ( k ) : 1+ 3 + 5 +K + ( 2k − 1) = k
We assume p(k) to be true ...
Write p(k+1), then start with p(k) and use math
manipulation to get p(k) to become p(k+1).
2
p ( k + 1) : 1+ 3 + 5 +K + ( 2k − 1) + ( 2k + 1) = ( k + 1)

31. Prove that the sum of the the first k odd positive
integers is k .
2
2
2. p ( k ) : 1+ 3 + 5 +K + ( 2k − 1) = k
We assume p(k) to be true ...
Write p(k+1), then start with p(k) and use math
manipulation to get p(k) to become p(k+1).
2
p ( k + 1) : 1+ 3 + 5 +K + ( 2k − 1) + ( 2k + 1) = ( k + 1)
2
1+ 3 + 5 +K + ( 2k − 1) = k

32. Prove that the sum of the the first k odd positive
integers is k .
2
2
2. p ( k ) : 1+ 3 + 5 +K + ( 2k − 1) = k
We assume p(k) to be true ...
Write p(k+1), then start with p(k) and use math
manipulation to get p(k) to become p(k+1).
2
p ( k + 1) : 1+ 3 + 5 +K + ( 2k − 1) + ( 2k + 1) = ( k + 1)
2
1+ 3 + 5 +K + ( 2k − 1) = k
2
1+ 3 + 5 +K + ( 2k − 1) + ( 2k + 1) = k + 2k + 1

33. Prove that the sum of the the first k odd positive
integers is k .
2
2
2. p ( k ) : 1+ 3 + 5 +K + ( 2k − 1) = k
We assume p(k) to be true ...
Write p(k+1), then start with p(k) and use math
manipulation to get p(k) to become p(k+1).
2
p ( k + 1) : 1+ 3 + 5 +K + ( 2k − 1) + ( 2k + 1) = ( k + 1)
2
1+ 3 + 5 +K + ( 2k − 1) = k
2
1+ 3 + 5 +K + ( 2k − 1) + ( 2k + 1) = k + 2k + 1
2
1+ 3 + 5 +K + ( 2k − 1) + ( 2k + 1) = ( k + 1)
and our proof is complete!

34. Page 859 #7 2 2
3 3 3 3 n ( n + 1)
1 + 2 + 3 +K + n =
4

35. Page 859 #7 2 2
3 3 3 3 n ( n + 1)
1 + 2 + 3 +K + n =
4
Proof:

36. Page 859 #7 2 2
3 3 3 n ( n + 1)
3
1 + 2 + 3 +K + n =
4
Proof:
2 2
31 (1+ 1)
p (1) : 1 =
4

37. Page 859 #7 2 2
3 3 3 n ( n + 1)3
1 + 2 + 3 +K + n =
4
Proof:
2 2
31 (1+ 1)
p (1) : 1 =
4
2
1( 2 )
1=
4

38. Page 859 #7 2 2
3 3 3 n ( n + 1)3
1 + 2 + 3 +K + n =
4
Proof:
2 2
31 (1+ 1)
p (1) : 1 =
4
2
1( 2 )
1=
4
4
1=
4

39. Page 859 #7 2 2
3 3 3 n ( n + 1)3
1 + 2 + 3 +K + n =
4
Proof:
2 2
1 (1+ 1)
3
p (1) : 1 =
4
2
1( 2 )
1=
4
4
1=
4
1 = 1 ∴ p (1) is true
(continued on next slide)

40. Page 859 #7 2 2
3 3 3 n ( n + 1)
3
1 + 2 + 3 +K + n =
4
2 2
3 3 3 3 n ( n + 1)
p ( n ) : 1 + 2 + 3 +K + n =
4

41. Page 859 #7 2 2
3 3 3 n ( n + 1)
3
1 + 2 + 3 +K + n =
4
2 2
3 3 3 n ( n + 1)
3
p ( n ) : 1 + 2 + 3 +K + n =
4
2 2
3 3 3 3
p ( n + 1) : 1 + 2 + 3 +K + n + ( n + 1)
3
=
( n + 1) ( n + 2 )
4

42. Page 859 #7 2 2
3 3 3 n ( n + 1)
3
1 + 2 + 3 +K + n =
4
2 2
3 3 3 n ( n + 1)
3
p ( n ) : 1 + 2 + 3 +K + n =
4
2 2
3 3 3 3
p ( n + 1) : 1 + 2 + 3 +K + n + ( n + 1)
3
=
( n + 1) ( n + 2 )
4

43. Page 859 #7 2 2
3 3 3 n ( n + 1)
3
1 + 2 + 3 +K + n =
4
2 2
3 3 3 n ( n + 1)
3
p ( n ) : 1 + 2 + 3 +K + n =
4
2 2
3 3 3 3
p ( n + 1) : 1 + 2 + 3 +K + n + ( n + 1)
3
=
( n + 1) ( n + 2 )
4
2 2
3 3 3 n ( n + 1)
3
1 + 2 + 3 +K + n =
4

44. Page 859 #7 2 2
3 3 3 n ( n + 1)
3
1 + 2 + 3 +K + n =
4
2 2
3 3 3n ( n + 1)
3
p ( n ) : 1 + 2 + 3 +K + n =
4
2 2
3 3 3
p ( n + 1) : 1 + 2 + 3 +K + n + ( n + 1)3 3
=
( n + 1) ( n + 2 )
4
2 2
3 3 3 n ( n + 1)
3
1 + 2 + 3 +K + n =
4
2 2
3 3 3 3 n ( n + 1)
3 3
1 + 2 + 3 +K + n + ( n + 1) = + ( n + 1)
4
(continued on next slide)

45. 2 2
3 3 3 3
p ( n + 1) : 1 + 2 + 3 +K + n + ( n + 1)
3
=
( n + 1) ( n + 2 )
4
2 2
3 3 3 3 n ( n + 1)
3 3
1 + 2 + 3 +K + n + ( n + 1) = + ( n + 1)
4

46. 2 2
3 3 3 3
p ( n + 1) : 1 + 2 + 3 +K + n + ( n + 1)
3
=
( n + 1) ( n + 2 )
4
2 2
3 3 3 3 n ( n + 1)
3 3
1 + 2 + 3 +K + n + ( n + 1) = + ( n + 1)
4
2 2 3
3 3 3 3 n ( n + 1) + 4 ( n + 1)
3
1 + 2 + 3 +K + n + ( n + 1) =
4

47. 2 2
3 3 3 3
p ( n + 1) : 1 + 2 + 3 +K + n + ( n + 1)
3
=
( n + 1) ( n + 2 )
4
2 2
3 3 3 3 n ( n + 1)
3 3
1 + 2 + 3 +K + n + ( n + 1) = + ( n + 1)
4
2 2 3
3 3 3 3 n ( n + 1) + 4 ( n + 1)
3
1 + 2 + 3 +K + n + ( n + 1) =
4
2
3 3 3 3
1 + 2 + 3 +K + n + ( n + 1) =
3 ( n + 1) (n 2
+ 4 ( n + 1))
4

48. 2 2
3 3 3 3
p ( n + 1) : 1 + 2 + 3 +K + n + ( n + 1)
3
=
( n + 1) ( n + 2 )
4
2 2
3 3 3 3 n ( n + 1)
3 3
1 + 2 + 3 +K + n + ( n + 1) = + ( n + 1)
4
2 2 3
3 3 3 3 n ( n + 1) + 4 ( n + 1)
3
1 + 2 + 3 +K + n + ( n + 1) =
4
2
3 3 3 3
1 + 2 + 3 +K + n + ( n + 1) =
3 ( n + 1) (n 2
+ 4 ( n + 1))
4
2
3 3 3 3
1 + 2 + 3 +K + n + ( n + 1) =
3 ( n + 1) (n 2
+ 4n + 4 )
4

49. 2 2
3 3 3 3
p ( n + 1) : 1 + 2 + 3 +K + n + ( n + 1)
3
=
( n + 1) ( n + 2 )
4
2 2
3 3 3 3 n ( n + 1)
3 3
1 + 2 + 3 +K + n + ( n + 1) = + ( n + 1)
4
2 2 3
3 3 3 3 n ( n + 1) + 4 ( n + 1)
3
1 + 2 + 3 +K + n + ( n + 1) =
4
2
3 3 3 3
1 + 2 + 3 +K + n + ( n + 1) =
3 ( n + 1) (n 2
+ 4 ( n + 1))
4
2
3 3 3 3
1 + 2 + 3 +K + n + ( n + 1) =
3 ( n + 1) (n 2
+ 4n + 4 )
4
2 2
3 ( n + 1) ( n + 2 )
QED
3 3 3 3
1 + 2 + 3 +K + n + ( n + 1) =
4

50. HW #9
“A leader is anyone who has two characteristics; first,
he is going somewhere; second, he is able to persuade
other people to go with him.” W. H. Cowley